THE p-ADIC VALUATION OF STIRLING NUMBERS - CiteSeerX

THE p-ADIC VALUATION OF STIRLING NUMBERS ANA BERRIZBEITIA, LUIS A. MEDINA, ALEXANDER C. MOLL, VICTOR H. MOLL, AND LAINE NOBLE Abstract. Let p > 2 be a prime. The p-adic valuation of Stirling numbers of the second kind is analyzed. Two types of tree diagrams that encode this information are introduced. Conditions that describe the infinite branching of these trees, similar to the case p = 2, are presented.

1. Introduction The Stirling numbers of second kind S(n, k), defined for n ∈ N and 0 ≤ k ≤ n count the number of ways to partition a set of n elements into exactly k nonempty subsets. They are explicitly given by   k−1 k 1 X (k − i)n , (−1)i (1.1) S(n, k) = i k! i=0 or, alternatively, by the recurrence (1.2)

S(n, k) = S(n − 1, k − 1) + kS(n − 1, k),

with the initial conditions S(0, 0) = 1 and S(n, 0) = 0 for n > 0. Divisibility properties of integer sequences are expressed in terms of p-adic valuations: given a prime p and a positive integer b, there exist unique integers a, n, with a not divisible by p and n ≥ 0, such that b = apn . The number n is called the p − adic valuation of b. We write n = νp (b). Thus, νp (b) is the highest power of p that divides b. The 2-adic valuation of the Stirling numbers was discussed in [1]. The main conjecture described there is that the partitions of the positive integers N in classes of the form (1.3)

Cm,j := {2m i + j : i ∈ N}

where the index i starts at the point where 2m i + j ≥ k, leads to a clear pattern for the 2-adic valuations of Stirling numbers. The class Cm,j is called constant if ν2 (Cm,j ) consists of a single value. This single value is called the constant of the class Cm,j . The parameter m in (1.3) is called the level of the class. The level are defined inductively as follows: assume that the (m − 1)-level has been defined and that it consists of the s classes (1.4)

Cm−1,i1 , Cm−1,i2 , . . . , Cm−1,is .

Date: August 31, 2009. 2000 Mathematics Subject Classification. Primary 11B50, Secondary 05A15. Key words and phrases. valuations, Stirling numbers. 1

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Each class Cm−1,ij splits into two classes modulo 2m , namely Cm,ij and Cm,ij +2m−1 . The m-level is formed by the non-constant classes modulo 2m . The main conjecture of [1] is now stated: Conjecture 1.1. Let k ∈ N be fixed. Then we conjecture that a) there exists a level m0 (k) and an integer µ(k) such that for any m ≥ m0 (k), the number of nonconstant classes of level m is µ(k), independently of m; b) moreover, for each m ≥ m0 (k), each of the µ(k) nonconstant classes splits into one constant and one nonconstant subclass. The latter generates the next level set. This conjecture was established in [1] only for the case k = 5. The proof makes strong use of the recurrence (1.2) and the fact that the 2-adic valuations for S(n, k), for 1 ≤ k ≤ 4 are easily determined. The goal of this paper is to describe a similar behavior for the p-adic valuations of S(n, k) for the case of p an odd prime. 2. Modular and p-adic trees Consider a sequence of positive integers a := {aj }, with aj dividing aj+1 . The tree associated to a starts with a root vertex and as first generation has a1 vertices representing all the residue classes modulo a1 . Each vertex, branches into the second generation into a2 /a1 siblings corresponding to residues modulo a2 . For instance, the vertex corresponding to 1 in the first generation branches into the vertices corresponding to the classes 1, 1 + a1 , 1 + 2a1 , . . . , 1 + (a2 /a1 − 1)a1 . The m-th generation of the tree contains am vertices that provide a partition of N into a collection of residue classes. We define two types of trees that encode the p-adic valuations of the Stirling numbers. In each case, the sequence aj described above corresponds to the period of the function S(n, k) modulo powers of the prime p. This periodicity is described first. Given a prime p and an integer m ∈ N, the sequence S(n, k) mod pm , for k fixed is periodic of period Lm := (p − 1)pm+α(p,k)−1 , where α(p, k) is defined by pα(p,k) < k ≤ pα(p,k)+1 . Observe that Lm divides Lm+1 with quotient p. For instance, the Stirling numbers S(n, 5) modulo 32 have period 18, the repeating block being (2.1)

{1, 6, 5, 6, 3, 0, 4, 6, 8, 6, 6, 0, 7, 6, 2, 6, 0, 0}.

The question of the minimal period of S(n, k) modulo pm has been discussed by Nijenhuis and Wilf [5] and Kwong [4]. This paper is not concerned with minimality and uses only the period Lm defined above. See [2], [3] for more information on this topic. Indexing. Fix a prime p and m ∈ N. The residue classes modulo Lm+1 can be written in the form (2.2)

n = i0 + i1 L 1 + i2 L2 + · · · + im Lm ,

where 0 ≤ i0 ≤ p − 2 = L1 − 1 and 0 ≤ ir ≤ p − 1 for 1 ≤ r ≤ m. The coefficients ir are uniquely determined by n.

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Definition 2.1. The set (2.3)

Dm,j := {n ∈ N : n ≡ j

mod Lm },

will be called a class at the m-th level. The tree associated to the sequence {Lj } is now modified by branching a subset of the nodes at the m-th level. Recall that each vertex at the m-th level corresponds to a class Dm,j . Two different branching criteria produce the two types of trees mentioned above. Note. We assume throughout that k < p. This has the advantage that the term k! in (1.1) is invertible modulo p, so it may be ignored in the question of divisibility of S(n, k) by p. This assumption also yields α(p, k) = 0 and the sequence S(n, k) mod pm is periodic of period Lm := (p − 1)pm−1 = ϕ(pm ), where ϕ is the Euler totient function. Modular trees. A branching vertex is one for which (2.4)

S(Dm,j , k) ≡ 0 mod pm .

The remaining vertices will be called terminal. The periodicity of S(n, k) modulo pm shows that (2.4) is independent of the element in the class Dm,j . Indeed, if n1 ≡ n2 mod p − 1, then (2.5)

S(n1 , k) ≡ S(n2 , k) mod p.

In particular, if S(n1 , k) 6≡ 0 mod p, then νp (S(n, k)) = 0 for all n ≡ n1 mod (p−1). On the other hand, if S(n1 , k) ≡ 0 mod p, then is not guaranteed that νp (S(n1 , k)) and νp (S(n2 , k)) are the same.

0

0

Figure 1. The first level for p = 5 and k = 3.

The first level consists of p − 1 vertices corresponding to the residue classes modulo L1 = p − 1. The branches that correspond to values with S(n, k) 6≡ 0 mod p are labeled by the valuation, namely 0, they are terminated. The remaining branches are not labeled and are split modulo (p − 1)p into the next level. The p edges coming out of a branching vertex i correspond to the p numbers nj := i + (p − 1)j with 0 ≤ j ≤ p − 1. All these indices are congruent modulo p − 1 and distinct modulo L2 := p(p − 1). Their corresponding Stirling numbers

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0

0

1 1 1 1

Figure 2. The first and second level for p = 5 and k = 3.

satisfy νp (S(nj , k)) ≥ 1. At this point, we consider the values of S(nj , k) modulo p2 . Using the fact that (2.6)

a ≡ b mod L2 implies S(a, k) ≡ S(b, k) mod p2 ,

we conclude that, if S(nj , k) 6≡ 0 mod p2 , then νp (S(nj , k)) = 1. Those vertices are labeled 1 and terminated. The remaining vertices, namely those for which S(nj , k) ≡ 0 mod p, are split again modulo p. For these classes we have that their valuations are at least 2. Example 2.2. Figures 1 and 2 illustrate the first two levels for p = 5 and k = 3. At the first level, the integers are divided in classes modulo L1 = 4 and the corresponding Stirling numbers are considered modulo p = 5. Figure 1 shows that two classes, those congruent to 0 and 3 modulo 4, have Stirling numbers not divisible by 5. Thus, ν5 (D1,0 ) = ν5 (D1,3 ) = 0. These branches are labeled by their valuation. The Stirling numbers corresponding to the two remaining classes do not have the same valuation. At this point one can only conclude that ν5 (n, 3) ≥ 1 for n ≡ 1 or 2 mod 4. For example, ν5 (S(13, 3)) =

ν5 (7508501) = 3

ν5 (S(17, 3)) =

ν5 (5652751651) = 2.

The numbers congruent to 1 modulo 4 split into classes congruent to 1, 5, 9, 13, 17 modulo L2 = 20 = 4 · 5. Similarly, the class 2 modulo 4 becomes the five classes 2, 6, 10, 14, 18 modulo 20. The construction of the tree continues by testing the Stirling numbers corresponding to these indices modulo 52 = 25. Lemma 2.3. If a vertex j appears at the level m, then νp (Dm,j ) ≥ m − 1. The modular tree has the advantage that it is easy to generate: at the m-th level the Stirling numbers are tested modulo pm . Moreover, if a branch terminates at level m, then the corresponding indices satisfy νp (n) = m. Experimentally it is found that they become very large. This is a disadvantage. An alternative to these trees is proposed next. p-adic trees. This second type of trees differ from the modular trees in the mechanism employed to terminate a branch. Recall that, at level m, each vertex corresponds to a class Dm,j . In this new type of trees, a vertex is declared terminal if the corresponding class has constant p-adic valuation; i.e. for n ∈ Dm,j , the valuation νp (S(n, k)) is independent of n. In practice, this is decided in an experimental

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manner. Given a parameter T∞ , the class Dm,j is declared constant if its first T∞ -values have the same valuation. Naturally a class is branched if it contains two elements with distinct valuation. The last section presents an algorithm where this experimental procedure is replaced by a rigorous procedure.

0

0 1111

22222

22222

22222

2222

Figure 3. The first three levels of the modular tree for p = 5 and k=3

0

22

3333

0

2

33

1111

33

2222

Figure 4. The first three levels of the p-adic tree for p = 5 and k = 3

Figures 3 and 4 present the first three levels of the modular and 5-adic trees for k = 3, respectively. Observe that in the modular trees there are vertices that branch to the third level only to discover that each of the siblings has the same constant valuation. This is the case for the second vertex at the second level. The p-adic tree detects this phenomena at the second level. This accounts for the fact that, in general, p-adic trees are smaller that the modular trees. The study of the p-adic valuations of Stirling numbers begins with an elementary criteria. Lemma 2.4. Let i be a residue class modulo L1 such that S(i, k) 6≡ 0 mod p. Then i is a terminal vertex and the class D1,i has p-adic valuation 0. Example 2.5. Take p = 7 and k = 4. The periodicity of the Stirling numbers shows that if i ≡ 0 mod L1 = 6, then S(i, 4) ≡ S(6, 4) mod 7. In this case S(6, 4) =

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65 ≡ 2 mod 7, thus S(6t, 4) ≡ 2 mod 7. Therefore the class D1,0 has constant 7-adic valuation 0 and the vertex 0 is terminal.

0

0

0

Figure 5. The first level for p = 7 and k = 4

Figure 5 shows that the vertices 0, 4 and 5 are terminal and the classes D1,0 , D1,4 and D1,5 have constant valuation 0. The remaining classes D1,1 , D1,2 and D1,3 are now tested further. The description is given for the class D1,1 , the others are treated using the same ideas. Details are given in section 5. For level m = 1 and if n ≡ 1 mod L1 = 6, we have (2.7)

S(1 + 6t, 4) ≡ S(L1 + 1, 4) = S(7, 4) mod 7.

The value S(7, 4) = 350, shows that S(7, 4) ≡ 0 mod 7. We conclude that the valuation ν7 (S(1 + 6t, 4)) ≥ 1. The question of whether the class D1,1 is constant is decided in an experimental manner. The required number of values to declare the branching of a class is usually small. For instance, for D1,1 , the value T∞ = 7 suffices in practice. Indeed, S(43, 4) = 72 × N , with N = 65790764819319273461750 6≡ 0 mod 7. Therefore, the class D1,1 is not constant. The value T∞ = 7 is also sufficient to determine that the two remaining classes, D1,2 and D1,3 are nonconstant. The analysis of the branched class D1,1 is now considered at level 2, that is, the sequence n = 1 + 6t is considered modulo L2 = (p − 1)p = 42. The class D1,1 splits into the seven classes (2.8)

D2,1 , D2,7 , D2,13 , D2,19 , D2,25 , D2,31 , D2,37 .

The branches of the tree depicted in Figure 6 show the branching from the nonconstant class D1,1 . Recall the indexing for residue classes modulo L2 = 42 in the form n = i0 + i1 L1 described in (2.2). The subtree depicted in Figure 6 corresponds to indices with i0 = 1 and the label in each branch is the residue of S(i0 + i1 L1 + tL2 , 4) modulo 72 . Note. Figure 6 shows that six of the seven classes in (2.8) are constant. This leaves only the leftmost vertex, with label i0 = 1, i1 = 0, for possible branching. This is the class (2.9)

D2,1 = {n ∈ N : n ≡ 1 mod 42} = {1 + 42t : t ∈ N}.

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D1,1

@1,0D

@7,7D

@13,14D

@19,21D

@25,28D

@31,35D

@37,42D

Figure 6. Branching into the second level

0

111111

00

111111

111111

Figure 7. The second level for p = 7 and k = 4

Then one checks that D2,1 branches to level 3 by computing the values of ν7 (S(1 + 42t, 4)) for 0 ≤ t ≤ T∞ . Once again, T∞ = 7 is sufficient to verify that D2,1 is a non-constant class. Note. In order to declare a class Dm,j constant, we compute the first T∞ values of νp (S(j + tLm , k)) for a precribed length T∞ . The class Dm,j is considered constant if these values are the same. Experiments have shown that a modest value T∞ suffices in each case. 3. Criteria for branching on modular trees The goal of this section is to discuss the branching mechanism on modular trees. The assumption k < p and the labeling described in (2.2) are imposed throughout. Recall that each vertex at level m represents a class Dm,j = {n ∈ N : n ≡ j mod Lm }. Using the value L1 = p − 1, it follows that level 1 is represented by

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the numbers {0, 1, 2, . . . , p − 2} and a vertex i0 branches to level 2 if (3.1)

S(i0 + tL1 , k) ≡ 0 mod p.

The terminal (= non-branching) vertices stop at level 1 and they are labeled by the valuation ( = 0). The branched vertices are now considered modulo L2 and they branch to level 3 if (3.2)

S(i0 + i1 L1 + tL2 , k) ≡ 0 mod p2 .

Continuing with this process, the vertices appearing at level m correspond to branches that have not been terminated up to this point. Therefore the corresponding classes satisfy νp (Dm,j ) ≥ m − 1. Lemma 3.1. Assume the a vertex i at level m satisfies S(i, k) 6≡ 0 mod pm .

(3.3)

Then νp (Dm,i ) = m − 1 and the vertex is terminal. The first result states that the left most vertex always terminates. Lemma 3.2. Assume n ≡ 0 mod L1 . Then S(n, k) 6≡ 0 mod p and νp (S(n, k)) = 0. Therefore the vertex D1,0 is terminal. Proof. Write n = i(p − 1) and then (3.4)

  k (−1)k X r k ri(p−1) . (−1) S(n, k) = k! r=1 r

Fermat’s little theorem shows that rp−1 ≡ 1 mod p, thus   k (−1)k X r k (3.5) S(n, k) ≡ (−1) mod p. k! r=1 r The result now follows from (3.6)

k X

  k = −1. (−1) r r=1 r

 Definition 3.3. Assume 1 ≤ r ≤ p − 1, so that rL1 ≡ 1 mod p. Define βr by the identity rL1 ≡ 1 + pβr mod p2 . The distribution of βr is complicated. Figure 8 shows the values of βr for 2 ≤ r ≤ 7918. The prime 7919 is the 1000-th prime. The expression βr is now employed to compute the residue of S(n, k) modulo p2 . Theorem 3.4. Let n ≡ i0 + i1 L1 mod L2 . Then # " k     k X (−1)k X i0 r k i0 r k r βr mod p2 . (−1) (3.7) S(n, k) ≡ r + i1 p (−1) r k! r r=1 r=1

p-ADIC VALUATION

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8000

6000

4000

2000

2000

4000

6000

8000

Figure 8. The values of βr for the prime p = 7919.

Proof. The relation rL2 ≡ 1 mod p2 yields   k (−1)k X r k S(n, k) ≡ ri0 ri1 L1 mod p2 . (−1) k! r=1 r

(3.8) Therefore S(n, k) ≡ ≡

≡

≡

  k (−1)k X r k (−1) ri0 ri1 L1 mod p2 k! r=1 r   k (−1)k X r k ri0 (1 + pβr )i1 mod p2 (−1) k! r=1 r   k (−1)k X k i0 r (1 + i1 pβr ) mod p2 (−1)r k! r=1 r     k k X k i0 k i0 (−1)k (−1)k X r βr mod p2 . r + i1 p (−1)r (−1)r k! r=1 k! r r r=1

This gives the result.



Corollary 3.5. Let n ≡ i0 mod L1 . Then (3.9)

  k (−1)k X r k S(n, k) ≡ ri0 mod p. (−1) k! r=1 r

Therefore the class D1,i0 branches to level 2 precisely when   k X r k ri0 ≡ 0 mod p. (3.10) (−1) r r=1

Corollary 3.6. Let n ≡ i0 mod L1 with 1 ≤ i0 ≤ k − 1. Then the class D1,i0 branches to level 2. Proof. The result follows from the previous corollary and   k X k i0 (3.11) (−1)r r = (−1)k k!S(i0 , k) = 0 r r=1

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in view of the assumption 1 ≤ i0 ≤ k − 1.



Corollary 3.7. Let D2,j be a class at level 2 with j ≡ i0 + i1 L1 mod L2 . Then D2,j branches to level 3 precisely when   k X r k ri0 ≡ 0 mod p2 (3.12) (−1) r r=1 and

(3.13)

i1

k X

(−1)r

r=1

  k i0 r βr ≡ 0 mod p. r

Corollary 3.8. Let D2,j be a class at level 2 with j ≡ i0 + i1 L1 mod L2 . Assume 1 ≤ i0 ≤ k − 1. Then D2,j branches to level 3 precisely when   k X k i0 (3.14) i1 (−1)r r βr ≡ 0 mod p. r r=1 In particular, the class corresponding to i1 = 0 always branches.

The remainder of this section addresses patterns of branching behavior that continue through increasing m-levels. Lemma 3.9. For given p and 1 ≤ r < p, m

rϕ(p

)

≡ 1 + βr pm mod pm+1

for all m, with βr independent of m. m−1

Proof. Suppose rϕ(p

)

m

rϕ(p

)

≡ 1 + βr pm−1 mod pm for some m. Then m−1

= ≡

)p rϕ(p (1 + βr pm−1 + apm )p mod pm+1

≡

1 + βr pm mod pm+1 .

This is the result.



Now consider a branching class Dm,j at level m. Its siblings consist of the p classes Dm+1,j+iLm with 0 ≤ i ≤ p − 1. The next result states a condition that guarantees the preservation of the pairwise incongruent property. In particular, from some point on, exactly one of them will branch. Theorem 3.10. Let m ≥ 1. Assume that for some branching class Dm,j all vertices are incongruent, that is, (3.15)

S(Dm+1,j+i1 Lm , k) 6≡ S(Dm+1,j+i2 Lm , k) mod pm+1

for 0 ≤ i1 < i2 ≤ p − 1. Then, every subsequent branching class DM,J with J ≡ j mod Lm and M ≥ m, (3.16)

S(DM+1,J+i1 LM , k) 6≡ S(DM+1,J+i2 LM , k) mod pM+1

for 0 ≤ i1 < i2 ≤ p − 1.

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Proof. Let d1 , d2 ∈ Dm,j such that d1 6≡ d2 mod Lm+1 . Index these values by d1 d2

= j + i1 Lm + x1 Lm+1 , = j + i2 Lm + x2 Lm+1 ,

with x1 , x2 ∈ Z. By assumption, S(d2 , k) − S(d1 , k) 6≡ 0 mod pm+1 . Then k

(−1) k! [S(d2 , k) − S(d1 , k)] ≡ ≡

≡

≡

k X

  k [rd2 − rd1 ] mod pm+1 (−1) r r=1   k X k j Lm i1 Lm i2 −i1 − 1] mod pm+1 r (r ) [(r ) (−1)r r r=1   k X r k rj (1 + βr pm )i1 [(1 + βr pm )i2 −i1 − 1] mod pm+1 (−1) r r=1   k X m r k (i2 − i1 )p rj βr mod pm+1 . (−1) r r=1 r

Observe that the hypothesis of pairwise incongruence imply that k X

  k j r βr 6≡ 0 mod p (−1) r r=1 r

and conversely. Since the Stirling numbers associated with the p vertices emanating from Dm,j are by hypothesis pairwise incongruent modulo pm+1 , exactly one of these vertices branches. Suppose that this branching vertex corresponds to the class Dm+1,j+i∗ Lm . Now write d1

= j + i∗ Lm + i∗1 Lm+1 + x1 Lm+2

d2

= j + i∗ Lm + i∗2 Lm+1 + x2 Lm+2

where again 0 ≤ i∗1 < i∗2 ≤ p − 1. To verify that the pairwise incongruence condition holds at this new level, it is required to show that S(d2 , k) − S(d1 , k) 6≡ 0 mod pm+2 .

(3.14)

Proceeding as before, we have k

(−1) k! [S(d2 , k) − S(d1 , k)] ≡ Expanding as before, it follows that

k X

   k  d2 (−1) r − r d1 r r=1

(−1)k k! [S(d2 , k) − S(d1 , k)] ≡ (i∗2 − i∗1 )pm+1

r

k X r=1

(−1)r

mod pm+2 .

  k j+i∗ Lm βr mod pm+2 . r r

The assumption (3.13) proves the pairwise incongruence of the Stirling numbers at level m + 2. 

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Corollary 3.11. Assume the class Dm,j has satisfies the pairwise incongruence condition at level m. Then it satisfies at level m + 1 if and only if   k X r k rj βr 6≡ 0 mod p. (3.13) (−1) r r=1 holds. Observe that this condition is independent of m.

Note. A branching class Dm,j satisfying condition (3.13) separates at level m + 1 into p − 1 terminal vertices (with Stirling numbers of constant p-adic valuation m) and exactly one branching vertex. Observe that this is exactly the analog of the main conjecture of [1]. Note. Let j be an index such that the hypothesis of Theorem 3.10 is satisfied for Dm,j at level m. Write (3.14)

j = i0 + i1 L1 + · · · + im−1 Lm−1 ,

and observe that rj = ri0

m−1 Y

rLt it ≡ ri0 mod p.

t=1

Thus, the condition (3.13) depends only on j modulo L1 . Corollary 3.12. Using the expansion (3.15)

j = i0 + i1 L1 + · · · + im−1 Lm−1 ,

the hypothesis of Theorem 3.10 is satisfied for Dm,j at level m if and only if it is satisfied for D1,i0 . Corollary 3.13. If (3.13) is not satisfied, that is,   k X k j r βr ≡ 0 mod p, (3.16) (−1)r r r=1

for j, then for d1 , d2 such that d1 ≡ d2 ≡ j mod L1 and d1 ≡ d2 mod Lm , S(d1 , k) ≡ S(d2 , k) mod pm+1 . Example 3.14. This example continues Example 2.2 for p = 5 and k = 3. Congruence (3.16) is satisfied for the class D1,1 , that is, for n ≡ 1 mod L1 = 4. Since 29 ≡ 49 ≡ 1 mod 4 and 29 ≡ 49 mod 20, Corollary 3.13 shows that S(29, 3) ≡ S(49, 3) mod 53 . This is depicted in Figure 9. Moreover, the class D2,9 corresponds to a terminal vertex: for all n such that n ≡ 9 mod 20, ν5 (S(n, 3)) = 2. 4. Refined branching criteria for modular trees In this section we describe a refinement of the criteria developed in the previous section. Examples illustrating the use of this criteria are given in Section 5. The goal is to provide a generalization of Theorem 3.10. We start by generalizing Definition 3.3. Definition 4.1. Fix a prime p. Given 1 ≤ r < p, define the sequence β1,r , β2,r , . . . , βω+1,r by (4.1)

ω+1

rϕ(p

)

≡ 1 + β1,r pω+1 + β2,r pω+2 + . . . + βω+1,r p2ω+1 mod p2ω+2 .

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D2,9

@0,25D

D3,9

@1,25D

D3,29

@2,25D

@3,25D

D3,49

@4,25D

D3,69

D3,89

Figure 9. Branching from the class D2,9 . S(9 + i1 · 20, k) ≡ S(9 + i2 · 20, k) mod 53 , for all i1 , i2 .

Next a generalization of Lemma 3.9. Lemma 4.2. The coefficients β1,r , β2,r , . . . , βω,r are independent on ω i.e. if m > ω then (4.2)

rϕ(m) ≡ 1 + β1,r pm + β2,r pm+1 + · · · + βω,r pω+m−1 mod pω+m

Proof. The proof is similar to the one for Lemma 3.9.



Definition 4.3. Define the condition   k X r k rj (β1,r + β2,r p + . . . + βω+1,r pω ) 6≡ 0 mod pω+1 (4.3) Tω+1,j := (−1) r r=1 for j modulo Lω+1 , ω ≥ 0. Observe that T1,j coincides with (3.13).

Proposition 4.4. Tω+1,j true (or false) implies Tω+1,j+im Lm true (or false) for all m ≥ ω + 1. Proof. For 1 ≤ r < p, rLm ≡ 1 mod pm , so rLm ≡ 1 mod pω+1 for m ≥ ω + 1.



Just as the condition T1,j served as a test for a particular branching behavior continuing through m-levels, the collection of conditions T1,j , T2,j , . . . , Tω,j now serves as a test. The implication of these conditions is presented in the following proposition and theorem. Proposition 4.5. Consider a branching class Dj,m . Suppose that Tj,m is false, then p2m divides S(j + iLm , k) for all i ∈ N precisely when   k X k j r ≡ 0 mod p2m . (−1)r (4.4) r r=1

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Proof. Note that (−1)k k!S(j + aLm , k) ≡

k X

  k j r mod p2m r r=1   k X
k j r β1,r pm + β2,r pm+1 + · · · + βn,r p2m−1 mod p2m . +a (−1)r r r=1 (−1)r

Since Tj,m is false, the second sum is congruent to zero modulo p2m . The result is established.  Theorem 4.6. Suppose that Dm,j is a branching class. If T1,j , T2,j , . . . , Tm−1,j are false and Tm,j is true, then all siblings of Dm,j at level m + 1 are incongruent modulo p2m . Moreover, all siblings of every subsequent branching class DM,J with J ≡ j mod Lm and M > m are incongruent modulo pm+M . Proof. The proof of this theorem is similar to the one of Theorem 3.10. Let d1 , d2 ∈ Dm,j such that d1 6≡ d2 mod Lm+1 . Index these values by d1

= j + i1 Lm + x1 Lm+1 ,

d2

= j + i2 Lm + x2 Lm+1

where i1 < i2 and x1 , x2 ∈ Z. Note that (4.5)

(−1)k k![S(d1 , k) − S(d2 , k)] ≡ (i2 − i1 )pm Tm,j mod p2m .

Here we are abusing the notation and letting   k X
k j r β1,r + β2,r p1 + · · · + βn,r pm−1 (4.6) Tm,j = (−1)r r r=1

rather than the test itself. By assumption Tm,j ≡ 6 0 mod pm , hence (4.5) is not 2m congruent to zero modulo p . The rest of the proof follows by induction as in Theorem 3.10.  Example 4.7. Theorem 4.6 describes the behavior of the class D2,11 for p = 5 and k = 4. T1,11 is false, but T2,11 is true. Then since 231 ≡ 391 ≡ 11 mod 20 and 231 6≡ 391 mod 100, S(231, 4) 6≡ S(391, 4) mod 54 . This is depicted in Figure 10. 5. The p-adic valuation of Stirling numbers In this section we present three illustrative examples of the algorithm developed in this paper. We present the data using p-adic trees, since they do not grow as fast as modular trees. A conjecture on the structure of these valuations is also presented. This conjecture extends the results presented in [1]. Recall the construction of trees described here: fix a prime p and an index k satisfying 0 ≤ k < p. The Stirling numbers S(n, k) are periodic modulo pm , of period Lm = (p − 1)pm−1 . The description of the patterns of the valuations νp (S(n, k)) is given in terms of the classes (5.1)

Dm,j = {n ∈ N : n ≡ j mod Lm }.

p-ADIC VALUATION

15

D2,11

0

D3,11

125 125 125 125 125

0

0

D3,31

0

0

D3,51

375 375 375 375 375

0

D3,71

0

0

0

0

250 250 250 250 250

D3,91

500 500 500 500 500

Figure 10. Branching from the class D2,11 , for p = 5 and k = 4.

Each class corresponds to a vertex of a tree and, for each m ∈ N, the collection of all classes Dm,j , is called the m-th level of the tree. The fundamental connection between classes and divisibility properties is given by (5.2)

a, b ∈ Dm,j if and only if S(a, k) ≡ S(b, k) mod pm .

Example 5.1. Let p = 7 and k = 3. Figure 11 presents the p-adic tree corresponding to these numbers.

0

000

111111 222222 333333

111111 222222

333333

Figure 11. The 7-adic tree of S(n, 3)

The first level of the tree associated with this case corresponds to the residues modulo L1 = 6. This gives the six classes D1,0 , D1,1 , D1,2 , D1,3 , D1,4 and D1,5 that form the first level. Lemma 3.2 guarantees that the class (5.3)

D1,0 = {n ∈ N : n ≡ 0 mod 6}

terminates and it has valuation 0. On the other hand, the classes (5.4) (5.5)

D1,3 D1,4

= =

{n ∈ N : n ≡ 3 mod 6} {n ∈ N : n ≡ 4 mod 6}

(5.6)

D1,5

=

{n ∈ N : n ≡ 5 mod 6}

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A. BERRIZBEITIA ET AL.

also terminate with valuation 0. The reason is simple: the class D1,3 contains 3 and S(3, 3) = 1. Periodicity of S(n, 3) implies that every index in D1,3 satisfies S(4t + 3, 3) ≡ 1 mod 7. Therefore ν5 (D1,3 ) = 0. Similarly, D1,4 contains 4 and S(4, 3) = 6 and D1,5 contains 5 and S(5, 3) = 25. The remaining two classes, D1,1 and D1,2 , satisfy S(D1,1 , k) ≡ S(D1,2 , k) ≡ 0 mod 7, hence testing is required. The D1,1 class has a simple branching structure. At level 2 D1,1 has seven siblings D2,1 , D2,7 , D2,13 , D2,19 , D2,25 , D2,31 and D2,37 . These classes are labeled by the index i0 + i1 L1 modulo L2 = 42. The label i0 = 1 and 0 ≤ i1 ≤ 6. A direct calculation shows that S(2 + 0 · 6 + tL2 , 3) ≡ 0 mod 72 S(2 + 1 · 6 + tL2 , 3) ≡ 7 mod 72 S(2 + 2 · 6 + tL2 , 3) ≡ 14 mod 72 S(2 + 3 · 6 + tL2 , 3) ≡ 21 mod 72 S(2 + 4 · 6 + tL2 , 3) ≡ 28 mod 72 . S(2 + 5 · 6 + tL2 , 3) ≡ 35 mod 72 S(2 + 6 · 6 + tL2 , 3) ≡ 42 mod 72 . Observe that these values are incongruent modulo 72 . An alternative way to see this is by noticing that condition (3.13) holds   3 X r 3 rβr ≡ 1 mod 7. (5.7) (−1) r r=1

Therefore, Theorem 3.10 states that this pattern of pairwise incongruence will continue through increasing m-levels, as illustrated in Figure 11. The class D1,2 also has a simple branching structure since a similar argument produces   3 X r 3 r2 βr ≡ 5 mod 7. (5.8) (−1) r r=1

It follows that each of the classes D1,1 and D1,2 split modulo L2 = 42 into six terminal classes, with constant ν7 (S(n, 3)) = 1, and one branching (nonconstant) class. This branching class will moreover split modulo L3 = 294 into six terminal classes with constant ν7 (S(n, 3)) = 2 and one branching class, and so on, at each m-level. This ends the characterization of this tree. Example 5.2. The next example considers the valuations ν5 (S(n, 4); that is, p = 5 and k = 4, see Figure 12. The first level is formed by the classes D1,0 , D1,1 , D1,2 and D1,3 . Lemma 3.2 states that the class D1,0 is terminal with valuation ν5 (D1,0 ) = 0. The remaining three classes are subject to the pre-test: is S(j + 4t, 4) ≡ 0 mod 5? If the answer is no, the vertex j is terminal and ν5 (D1,j ) = 0. From j ≤ k − 1 = 3, Theorem 3.6 shows that S(j + 4t, 4) ≡ 0 mod 5 for 1 ≤ j ≤ 3 and t ∈ N. In order to decide if these vertices branch to level 3, we ask the question: does T1,j hold?, that is, is   4 X 4 j (5.9) T1,j := (−1)r r β1,r 6≡ 0 mod 5 ? r r=1

p-ADIC VALUATION

17

0 1111 2222 3333 4444

1111 2222

3333 4444

2 3333

4444 5555

22 33 33 44 44 55 55

Figure 12. The 5-adic tree of S(n, 4)

Recall that βr is defined in (3.3). A direct calculation shows that (5.10)

β1,1 = 0, β1,2 = 3, β1,3 = 1, β1,4 = 1

and this gives that T1,1 and T1,2 are true and T1,3 is false. As in the case of Example 5.1, it follows that each of the classes D2,1 and D2,2 split modulo 20 into four terminal classes, with constant ν5 (S(n, 4)) = 1, and one branching (nonconstant) class. This branching class will moreover split modulo 100 into four terminal classes with constant ν5 (S(n, 4)) = 2 and one branching class, and so on, at each m-level. The class D1,3 corresponding to the vertex j = 3 requires further testing. The class {n ∈ N : n ≡ 3 mod 4} at the second level splits into five classes modulo 20: j = 3, 7, 11, 15, 19. Since   k X r k r3 ≡ 0 mod 52 , (−1) (5.11) r r=1

then Proposition 4.5 implies that 52 divides S(D2,j , k) for j = 3, 7, 11, 15, 19. Hence, for each of these, the following test is performed: first, is S(j + 20t, 4) ≡ 0 mod 53 ? If not, then the vertex j is terminal, with ν5 (S(n, 4)) = 2 for every n ≡ j mod 20. Then, does T2,j hold? The only j for which S(j +20t, 4) ≡ 0 mod 53 are 3 and 11, so it is only necessary to test T2,3 and T2,11 . Recall that β2,r is defined by (5.12)

r20 ≡ 1 + 52 β1,r + 53 β2,r mod 54

so β2,r , 1 ≤ r ≤ 4, is determined by the following calculations: 1 ≡ 20 2 ≡

1 + 52 · 0 + 53 · 0 mod 54 1 + 52 · 3 + 53 · 3 mod 54

320 420

1 + 52 · 1 + 53 · 0 mod 54 1 + 52 · 1 + 53 · 2 mod 54

≡ ≡

Thus we have β2,1 = 0, β2,2 = 3, β2,3 = 0, β2,4 = 2.

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A. BERRIZBEITIA ET AL.

Then       4 3 4 3 4 3 2 (3 + 3 · 5) − 3 + 4 (1 + 2 · 5) = 1460 6≡ 0 mod 52 2 3 4

(5.13)

implies T2,3 true, and       4 11 4 11 4 11 (5.14) 2 (3 + 3 · 5) − 3 + 4 (1 + 2 · 5) = 45649940 6≡ 0 mod 52 2 3 4 implies T2,11 true. So per Theorem 4.6, each of the classes j = 3, 11 at the second level splits modulo 100 into four terminal classes with ν5 (S(n, 4)) = 3 and one branching class. This branching class will moreover split modulo 500 into four terminal classes with constant ν5 (S(n, 4)) = 4 and one branching class, and so on, at each m-level. This completes our characterization of this tree. Example 5.3. The last example described here is p = 11 and k = 4, see Figure 13.

000000

0

1111 1 11111

2222 2 22222

3333 3 33333

4444 4 44444

1111 1 11111

2222 2 22222

444444444

2222 2 22222

3333 3 33333

4444 4 44444

5555 5 55555

6666 6 66666

55

66666 6 6

55 5 55555

666

Figure 13. The 11-adic tree of S(n, 4)

As in the previous two examples, the first level of this tree is formed by the classes D1,j , j = 0, 1, 2, · · · , 9. The classes D1,i , i = 0, 4, 5, 6, 7, 8, 9 are terminal with valuation ν5 (D1,i ) = 0. The classes D1,j for each 1 ≤ j ≤ 3 require test. Using Definition 3.3 and Example 5.2, the reader can verify that in this case we have β1,1 = 0, β1,2 = 5, β1,3 = 0, and β1,4 = 10. This information yields T1,1 true, T1,2 true, and T1,3 false. From here we gather that each of the vertices j = 1, 2 splits modulo 110 into ten terminal classes, with constant ν11 (S(n, 4)) = 1, and one branching (nonconstant) class. This branching class will moreover split modulo 1210 into ten terminal classes with constant ν11 (S(n, 4)) = 2 and one branching class, and so on, at each m-level. Since T1,3 is false, then the vertex j = 3 requires further testing. We proceed as in Example 5.2. At the second level, this vertex splits into eleven classes modulo 110. These are j = 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 103.

p-ADIC VALUATION

19

Note that   4 X 4 3 r ≡ 0 mod 112 , (−1)r r r=1

(5.15)

therefore by Proposition 4.5 we know that 112 divides S(D2,j , k) for each of these j. The following test is performed at each vertex: first, is S(j + 110t, 4) ≡ 0 mod 113 ? If not, then the vertex j is terminal, with ν11 (S(n, 4)) = 2 for every n ≡ j mod 110. Then, does T2,j hold? The only j for which S(j + 110t, 4) ≡ 0 mod 113 is j = 3, so it is only necessary to test T2,3 . Recall that β2,r is defined by (5.16)

r110 ≡ 1 + 112 β1,r + 113 β2,r mod 114 .

A direct calculation gives β2,1 = 0, β2,2 = 1, β2,3 = 4, β2,4 = 2. Then       4 3 4 3 4 3 2 (5 + 1 · 11) − 3 (0 + 4 · 11) + 4 (10 + 2 · 11) = −1936 ≡ 0 mod 112 2 3 4

and so T2,3 is false. This implies that further testing is required and so we still need to go another level down. The class {n ∈ N : n ≡ 3 mod 110} splits into eleven classes modulo 1210: j = 3, 113, 223, 333, 443, 553, 663, 773, 883, 993, 1103. Since (5.17)

k X

  k 3 r ≡ 0 mod 114 , (−1) r r=1 r

then 114 divides S(D3,j ) for each of these j. Once again, for each of these classes, the following two test are performed: first, is S(j + 13310t, 4) ≡ 0 mod 115 ? If not, then the vertex j is terminal, with ν11 (S(n, 4)) = 4 for every n ≡ j mod 1210. Then, does T3,j hold? The only j for which S(j + 1210t, 4) ≡ 0 mod 115 are 3 and 113. The calculation (5.18)

β3,1 = 0, β3,2 = 3, β3,3 = 0, β3,4 = 6

implies T3,3 and T3,113 are true. So by Theorem 4.6, each of the classes j = 3, 113 splits modulo 13310 into ten terminal classes with ν(S(n, 4)) = 5 and one branching class, and so on, through m-levels. This completes our characterization of this tree. This process is a systematic and very easily automated way to summarize the structure of the p-adic valuation of S(n, k) for fixed k < p. We conclude with an extension of Conjecture 1.1 for any odd prime. Conjecture 5.4. Fix the index k. Given a prime p, its is conjectured that a) there exists a level m0,p (k) and an integer µp (k) such that for any m ≥ m0,p (k), the number of non-constant classes at the level m is µp (k), independently of m; b) moreover, for each m ≥ m0,p (k), each of the µp (k) nonconstant classes splits into p − 1 constant and one nonconstant subclass. The latter generates the next level set.

20

A. BERRIZBEITIA ET AL.

Supporting software. This article is accompanied by the Mathematica package StirlingTrees.m available from the webpages: http://www.math.rutgers.edu/~lmedina/ and http://www.math.tulane.edu/~vhm/ Acknowledgements. This project started at MSRI-UP during the summer of 2008. The work was supported by National Security Agency (NSA) grant H9823008-1-0063, the National Science Foundation (NSF) grant 0754872, a donation from the Gauss Research Foundation in Puerto Rico, and The Mathematical Sciences Research Institute (MSRI). The fourth author acknowledges the partial support of NSF-DMS 0713836. References [1] T. Amdeberhan, D. Manna, and V. Moll. The 2-adic valuation of Stirling numbers. Experimental Mathematics, 17:69–82, 2008. [2] H. W. Becker and J. Riordan. The arithmetic of Bell and Stirling numbers. Amer. J. Math., 70:385–394, 1948. [3] L. Carlitz. Congruences for generalized Bell and Stirling numbers. Duke Math. J., 22:193–205, 1955. [4] Y. H. H. Kwong. Minimum periods of S(n, k) modulo M. The Fibonacci Quarterly, 27(3):217– 221, 1989. [5] A. Nijenhuis and H. S. Wilf. Periodicities of partition functions and Stirling numbers. J. Number Theory, 25:308–312, 1987. University of Wisconsin, Madison, WI 53706-1388 E-mail address: [email protected] Department of Mathematics, Rutgers University, Piscataway, NJ 08854 and Department of Mathematics, University of Puerto Rico, San Juan, PR 00931 E-mail address: [email protected] Columbia University, New York, NY 10027 E-mail address: [email protected] Department of Mathematics, Tulane University, New Orleans, LA 70118 E-mail address: [email protected] Ohio State University, Columbus, OH 43210 E-mail address: [email protected]